Captain Michael has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Stephanie and her merciless band of thieves. The Captain has probability $\dfrac{1}{2}$ of hitting the pirate ship, if his ship hasn't already been hit. If it has been hit, he will always miss. The pirate has probability $\dfrac{1}{6}$ of hitting the Captain's ship, if her ship hasn't already been hit. If it has been hit, she will always miss as well. If the Captain shoots first, what is the probability that the Captain hits the pirate ship, but the pirate misses?
Answer: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is the Captain hitting the pirate ship and event B is the pirate missing the Captain's ship. The Captain fires first, so his ship can't be sunk before he fires his cannons. So, the probability of the Captain hitting the pirate ship is $\dfrac{1}{2}$ If the Captain hit the pirate ship, the pirate has no chance of firing back. So, the probability of the pirate missing the Captain's ship given the Captain hitting the pirate ship is $1$ The probability that the Captain hits the pirate ship, but the pirate misses is then the probability of the Captain hitting the pirate ship times the probability of the pirate missing the Captain's ship given the Captain hitting the pirate ship. This is $\dfrac{1}{2} \cdot 1 = \dfrac{1}{2}$